Problem: Let $f$ be a transformation from $\mathbb{R}^3$ to $\mathbb{R}^3$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} 1 & 0 & 0 \\ \\ y & x & 0 \\ \\ yz & xz & xy \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $(2, 0, 3)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely
Explanation: The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} 1 & 0 & 0 \\ \\ y & x & 0 \\ \\ yz & xz & xy \end{bmatrix} \right) \\ \\ &= -1 (x \cdot xy - 0 \cdot xz) \\ \\ &- 0 (y \cdot xy - 0 \cdot yz) \\ \\ &+ (-1)(y \cdot xy - x \cdot yz) \\ \\ &= x^2y \end{aligned}$ If we evaluate $|J(f)|$ at $(2, 0, 3)$, we get $0$. Because the Jacobian determinant here is equal to $0$, we can conclude that $f$ will infinitely contract the space around $(2, 0, 3)$. To recap, the Jacobian determinant of $f$ is $x^2y$, and $f$ will infinitely contract the space around the point $(2, 0, 3)$.